Respuesta :
Answer: Mass of nitrous oxide used is 264 grams and mass of oxygen gas used is 96 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
- For a: Combustion of ethylene in nitrous oxide
The chemical equation for the combustion of ethylene in nitrous oxide follows the equation:
[tex]C_2H_4+N_2O\rightarrow 2CO_2+2H_2O+6N_2[/tex]
By stoichiometry of the reaction;
6 moles of nitrous oxide are required for the complete combustion of ethylene molecule.
Calculating the mass of nitrous oxide using equation 1:
Molar mass of nitrous oxide = 44 g/mol
Moles of nitrous oxide = 6 moles
Putting values in equation 1, we get:
[tex]6mol=\frac{\text{Mass of nitrous oxide}}{44g/mol}\\\\\text{Mass of nitrous oxide}=264g[/tex]
Thus, 264 grams of nitrous oxide are required for the complete combustion of ethylene.
- For b: Combustion of ethylene in air
The chemical equation for the combustion of ethylene in air follows the equation:
[tex]C_2H_4+3O_2\rightarrow 2CO_2+2H_2O[/tex]
By stoichiometry of the reaction;
3 moles of oxygen are required for the complete combustion of ethylene molecule.
Calculating the mass of oxygen using equation 1:
Molar mass of oxygen = 32 g/mol
Moles of oxygen = 3 moles
Putting values in equation 1, we get:
[tex]3mol=\frac{\text{Mass of oxygen}}{32g/mol}\\\\\text{Mass of oxygen}=96g[/tex]
Thus, 96 grams of oxygen are required for the complete combustion of ethylene.
